2020 AMC 10B Problem 24

The right side is an integer, so let $k = \lfloor \sqrt n \rfloor = \dfrac{n + 1000}{70}$. Then $n = 70k - 1000$, and the floor condition $k \le \sqrt n < k + 1$ becomes
$$k^2 \le 70k - 1000 < (k+1)^2.$$

Left inequality: $k^2 - 70k + 1000 \le 0$. The roots of $k^2 - 70k + 1000$ are $k = \frac{70 \pm \sqrt{4900 - 4000}}{2} = \frac{70 \pm 30}{2} = 20, 50$, so $20 \le k \le 50$.

Right inequality: $70k - 1000 < k^2 + 2k + 1$, i.e. $k^2 - 68k + 1001 > 0$. The roots are $k = \frac{68 \pm \sqrt{4624 - 4004}}{2} = 34 \pm \frac{\sqrt{620}}{2}$. Since $\sqrt{620} \approx 24.9$, the roots are about $21.55$ and $46.45$, so we need $k \le 21$ or $k \ge 47$.

Combining: $k \in \{20, 21, 47, 48, 49, 50\}$, six values. Each gives a positive $n = 70k - 1000$ (namely $400, 470, 2290, 2360, 2430, 2500$), and distinct $k$ give distinct $n$.

Spot check: $k = 21$ gives $n = 470$, $\sqrt{470} \approx 21.7$, floor $21$. Good. $k = 22$ gives $n = 540$, $\sqrt{540} \approx 23.2$, floor $23 \ne 22$. Fails, as predicted.

The answer is $\boxed{\textbf{(C)}\ 6}$.

A floor equation is really two inequalities. Naming the integer value $k$ converts the problem to one variable, and $n$ becomes a linear expression in $k$ while $\sqrt n$ becomes a quadratic condition. The line $70k - 1000$ and the parabolas $k^2$ and $(k+1)^2$ cross twice each, so the solutions live in two short intervals at the ends, not one long block. Always solve both inequalities; the answer choices $30$ and $32$ exist to catch students who stop after the first.

• Solving only k^2 <= n (getting 20 <= k <= 50, i.e. 31 values) and forgetting the upper bound n < (k+1)^2.
• Misjudging the boundary $k = 21$ or $k = 47$ by rounding $\sqrt{620}$ poorly; check the endpoints numerically.